Set h to the identity map, and an integral extension S/R becomes an integral homomorphism.
We already know that h induces a continuous map from spec S into spec R. When h is integral, this map becomes bicontinuous.
First assume h embeds R into S, and let vF be closed in spec S. Let E be the intersection of F and R. If every prime P in R containing E comes from at least one prime Q in S containing F, then the closed set vF maps to the closed set vE, and the function becomes bicontinuous. Let's see if we can lift P up to Q containing F.
Review the procedure for lifting P up to Q. We localized about P, then selected any maximal ideal in SP. But we could start with any proper ideal, such as FP, and raise this up to a maximal ideal. This pulls back to a prime ideal Q in S, containing F, and lying over P. Each P is covered, vF maps to vE, and the function is bicontinuous.
Next let h map R into S. This is a composition of two maps, from R onto h(R), and the embedding of h(R) into S. The latter induces a bicontinuous function from spec S into spec h(R), as shown above. Since prime ideals correspond under a ring homomorphism, h also induces a bicontinuous function. Combine these, and h(R) into S induces a bicontinuous function from spec S into spec R.
Let S T and M be ring extensions of R, and let h(S) be an integral homomorphism into T.
Tensor S T and h with M. Recall that S×M becomes a ring and an R algebra, and multiplication in this ring is performed per component. Also, the homomorphism from S×M into T×M maps xy to h(x)y. The image of x1x2y1y2 is h(x1x2)y1y2, or h(x1)h(x2)y1y2, or h(x1)y1 times h(x2)y2. The induced map is an R homomorphism, an S homomorphism, an M homomorphism, and a ring homomorphism. We want to show this is an integral homomorphism.
Let xy be a pair generator in T×M, and let p(x) be the monic polynomial that proves x is integral over S. The lead coefficient is 1; pair this with 1 to get 1 cross 1 in S×M. This is of course the identity element in the ring S×M. Pair the next coefficient with y, and the next one with y2, and the next one with y3, and so on. These are all elements of S×M. Evaluate this polynomial at xy, and pull out a common factor of yn. this gives p(x) cross yn, and since the former is 0 in T, the latter is 0 in T×M. Therefore xy is integral, and the induced ring homomorphism from S×M into T×M is integral.
If M is a fraction ring of R, tensoring with M implements a form of localization. Thus localization is a special case of this theorem. If P is a prime ideal of R, h becomes an integral ring homomorphism from SP into TP.
First, S×T is an R algebra. Let K be the tensor product of f(R) and g(R), and S×T becomes a ring extension of K, using the generators of S, as a ring extension of f(R), cross the generators of T, as a ring extension of g(R). Look here for a more complete characterization of S×T.
Let xy be an element in S×T. Let p(x) be a monic polynomial with coefficients in f(R), and let q(y) be a monic polynomial with coefficients in g(R). Raise p and q to appropriate powers, so that both have the same degree. Both polynomials remain monic, and x is still a root of p, and y is still a root of q.
Build a new polynomial w by merging the coefficients of p and q. The ith coefficient of w is the ith coefficient of p cross the ith coefficient of q, which represents an element of K. Thus w is monic with coefficients in K. Build a combined homomorphism from R into K, via f cross g. Each u cross v in f(R) cross g(R) can be comverted into uf(g-1(v)) cross 1. A linear combination of these images becomes something in f(R) cross 1. In other words, f cross g maps R onto K, and K is the image of R. Thus S×T is a K algebra, and an R algebra via f cross g. Furthermore, w is monic over K. We only need show xy is a root of w, and S×T becomes an integral R algebra.
Since multiplication takes place per component, xy times xy is x2y2. The jth term of w can be rewritten ajxjbjyj. Addition also takes place per component, so when the terms of w(xy) are added up, the result is p(x) cross q(y). This is 0 cross 0, or 0. Therefore xy is a root of w, and the tensor product of integral algebras is integral.